maverick280857
Apr18-05, 01:00 AM
Hi
This is a fairly straightforward problem but I want to do it using calculus. Here goes:
Prove that \sin 1 > \cos(\sin 1).
This is what I've done (I've hardly done much):
Let f(x) = \sin(\cos(\sin x) - \cos(\sin(\cos(x))). I also have to show that f(x) = 0 has exactly one solution in [0,\pi/2]. So anyway, f'(x) < 0 for all x in this interval which in particular means that f(0) > f(\pi/2). But the first part requires showing that f(0) > 0. Can this be done by a purely calculus-based argument? It is easily done if we bend the inequality a bit and observe that it is true if \pi/2 -1 <\sin 1.
Thanks and cheers
Vivek
This is a fairly straightforward problem but I want to do it using calculus. Here goes:
Prove that \sin 1 > \cos(\sin 1).
This is what I've done (I've hardly done much):
Let f(x) = \sin(\cos(\sin x) - \cos(\sin(\cos(x))). I also have to show that f(x) = 0 has exactly one solution in [0,\pi/2]. So anyway, f'(x) < 0 for all x in this interval which in particular means that f(0) > f(\pi/2). But the first part requires showing that f(0) > 0. Can this be done by a purely calculus-based argument? It is easily done if we bend the inequality a bit and observe that it is true if \pi/2 -1 <\sin 1.
Thanks and cheers
Vivek